Isolated singularity

A function $f$ has an isolated singularity at $z_0$ if $f$ is analytic on a punctured disk $0<|z-z_0|<r$ but not at $z_0$ itself — either $f$ is undefined at $z_0$ or fails to be analytic there.

Isolated means: there is a neighborhood of $z_0$ in which $z_0$ is the only point where $f$ misbehaves. Functions like $\sin (1/(1-z))$ near $z=1$ have non-isolated singularities; those are rarer and need different tools.

Three types

Distinguished by the principal part of the Laurent series at $z_0$:

Removable singularity

The Laurent series has no negative terms — it’s actually a Taylor series. Equivalently:

• ${\lim }_{z\to z_0}f(z)$ exists and is finite.
• $f$ is bounded near $z_0$.

The singularity can be “removed” by defining $f(z_0)={\lim }_{z\to z_0}f(z)$, making $f$ analytic at $z_0$.

Test: ${\lim }_{z\to z_0}f(z)$ exists and is finite. Or: $f$ is bounded near $z_0$.

Example. $f(z)=\sin z/z$ at $z=0$. Laurent / Taylor: $1-z^2/3!+z^4/5!-\cdots$ — no negative terms. Limit is $1$. Removable.

Pole of order $m$

The Laurent series has finitely many negative terms, with the most negative being $a_{-m}/(z-z_0{)}^m$, $a_{-m}\ne 0$. Equivalently:

• ${\lim }_{z\to z_0}(z-z_0{)}^{m}f(z)$ exists and is nonzero.

Test: $f$ blows up like $1/(z-z_0{)}^m$ at $z_0$.

Examples.

• $1/(z-z_0)$ has a simple pole (order 1) at $z_0$.
• $1/(z-z_0{)}^3$ has a pole of order 3.
• $1/[(z-1)(z-2{)}^2]$ has a simple pole at $z=1$ and a pole of order 2 at $z=2$.
• $\cos (z)/z$ has a simple pole at $z=0$ (since $\cos$ is analytic and nonzero at $0$).

Essential singularity

The Laurent series has infinitely many negative terms. The function does crazy things near $z_0$.

Picard’s theorem (deeper): in every neighborhood of an essential singularity, $f$ takes every complex value with at most one exception, infinitely many times.

Examples. $e^{1/z}$, $\sin (1/z)$, $\cos (1/z)$ — all at $z_0=0$.

For most practical problems, essential singularities don’t come up. Poles and removable singularities cover the standard cases.

Recognizing the type quickly

• If $f$ has a known formula and you can compute the limit at $z_0$: limit exists $\Rightarrow$ removable; limit is infinite $\Rightarrow$ pole or essential.
• For pole vs. essential: multiply by $(z-z_0{)}^m$ for increasing $m$ and check if the limit becomes finite (pole) or never finite (essential).
• Quick: a rational function $p(z)/q(z)$ in lowest terms has poles at zeros of $q$, of order equal to the multiplicity of that zero in $q$. Nothing essential.
• Functions involving $e^{1/g(z)}$, $\sin (1/g(z))$, etc., where $g(z)\to 0$ have essential singularities there.

Residues

The behavior near a singularity is encoded in the residue — the coefficient $a_{-1}$ of $1/(z-z_0)$ in the Laurent series. Different singularity types have different formulas:

• Removable: residue is $0$.
• Simple pole: $\mathrm{Res}(f,z_0)={\lim }_{z\to z_0}(z-z_0)f(z)$.
• Pole of order $m$: $\mathrm{Res}(f,z_0)=\frac{1}{(m-1)!}{\lim }_{z\to z_0}\frac{d^{m-1}}{d{z}^{m-1}}[(z-z_0{)}^{m}f(z)]$.
• Essential: read $a_{-1}$ off the Laurent expansion directly; no formula.

See Residue (complex analysis) for worked computations.

In context

The classification matters because:

• The Residue theorem requires evaluating residues; different types need different formulas.
• Removable singularities contribute zero to closed-contour integrals.
• Poles contribute $2\pi i\cdot \mathrm{Res}$ each.
• Essential singularities also contribute $2\pi i\cdot \mathrm{Res}$, but the residue must be extracted from the Laurent series by hand.